Number of solutions of a pair · 3 marks

What if one of a₂, b₂, c₂ is zero?

Answer: Then the ratio form cannot be used (we cannot divide by 0), but the idea still works in the form a₁ = ka₂, b₁ = kb₂, c₁ = kc₂ (k ≠ 0). So if c₂ = 0, there are infinitely many solutions only if c₁ = 0 too and the other coefficients are proportional; similarly for a₂ or b₂.

Step-by-step solution

Given: The pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0; a₁a₂ = b₁b₂ = c₁c₂ ⇒ infinitely many solutions

Idea: The ratio test is shorthand for “one equation is k times the other”. When a denominator is 0, go back to that statement: a zero coefficient in one equation must be matched by a zero in the same place in the other.

The real condition is that one equation is a non-zero multiple of the other: a₁ = ka₂, b₁ = kb₂, c₁ = kc₂. If, say, c₂ = 0, the fraction c₁c₂ is undefined, but c₁ = k × 0 just says c₁ = 0.1 mark
Example (c₂ = 0): 2x + 3y = 0 and 4x + 6y = 0. Here c₁ = c₂ = 0 and 24 = 36, so they are the same line: infinitely many solutions. But 2x + 3y = 1 and 4x + 6y = 0 (c₁ ≠ 0) are parallel: no solution.1 mark
Example (a₂ = 0): y = 2 written as 0x + y − 2 = 0, paired with x + y − 5 = 0. Since a₁ = 1 ≠ 0 = k × 0, they are not multiples; the lines cross at the single point (3, 2).½ mark
A safe way to avoid zero denominators is to cross-multiply: compare a₁b₂ with a₂b₁ (and b₁c₂ with b₂c₁). If a₁b₂ ≠ a₂b₁ there is a unique solution; if they are equal, look at the constants.½ mark

Use the multiple form instead of ratios: a₁ = ka₂, b₁ = kb₂, c₁ = kc₂ (k ≠ 0). A zero in one equation must be matched by a zero in the same place in the other for infinitely many solutions; cross-multiplying (a₁b₂ vs a₂b₁) avoids dividing by zero.

Check: (3, 2): 3 + 2 = 5 ✓ and y = 2 ✓. For 2x + 3y = 0 & 4x + 6y = 0: (3, −2) satisfies both ✓.

Answer to write in the exam

Condition: a₁ = ka₂, b₁ = kb₂, c₁ = kc₂ (k ≠ 0)

If c₂ = 0 ⇒ need c₁ = 0 (ratio c₁c₂ not usable)

e.g. 2x + 3y = 0 and 4x + 6y = 0 ⇒ infinitely many solutions

e.g. x + y = 5 and y = 2 ⇒ not multiples ⇒ unique solution (3, 2)

∴ Use a₁b₂ = a₂b₁ etc. instead of ratios when a coefficient is 0

Common mistakes that cost marks

Writing c₁0 and treating it as 0 or as infinity. Division by 0 is undefined; use the multiple form.
Assuming any pair with a zero coefficient has no solution.

How this can come in the exam

Short answer (2 marks)

Do 3x − 2y = 0 and 6x − 4y = 0 have infinitely many solutions? Explain without dividing by zero.

Show answer

6x − 4y = 2(3x − 2y) and the constants are both 0 = 2 × 0 (1 mark). So one is 2 × the other: same line, infinitely many solutions (1 mark).

Try one yourself

How many solutions do x = 4 and 2x + 0y = 8 have?